Question

If $e^{{\log }_e\left(1+x+x^2+......\right)}$ , then $\frac{dy}{dx}$is equal to?

• A. $\frac{1}{{(1+x)}^2}$
• B. $\frac{1}{{(1-x)}^2}$
• C. $\frac{-1}{{(1+x)}^2}$
• D. $\frac{-1}{{(1-x)}^2}$

Answer 1

The correct option is B

$\frac{1}{{(1-x)}^2}$

Explanation for the correct option:

Find the value of $\frac{dy}{dx}$ :

Given,

$e^{{\log }_e\left(1+x+x^2+......\right)}$

Now, evaluate the given function,

$y=1+x+x^2+\dots$ because $e^{\log x}=x$

$y=\frac{1}{1-x}\left[\because S_{\infty }=\frac{a}{1-r}\right]$ Infinite GP with $a=1$ and $r=x$

Now, differentiate the obtained function with respect to $x$

Then,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{(1-x)0-1(-1)}{{(1-x)}^2}\\ & =& \frac{1}{{(1-x)}^2}\end{array}$

Hence, the correct option is B.